Given that `E_(Ni^2+|Ni)^0= -0.25V and E_(Cu^2+|Cu)^0= +0.34V`. From this that suggest whether a solution of 1 (M) `CuSO_4` be stored in a nickel vessel or not.

Calculate the electrode potential given that E° (Zn^(2+) // Zn) = -0.76V , when a zinc rod is dipped in 0.1 M solution of ZnSO_(4) .The salt is 95% dissociated at this solution at 298K.

Two metals A and B have E_(RP)^0 = -0.76V and E_(RP)^0 =+0.80V respectively. Which metal will diplace H_2 from dilute H_2SO_4 ?

E^@ Fe //Fe^(2+) = 0.44 whereas E^@ Cu // Cu^2+ = -0.32 v Then

For the electrochemical cell, M[M^+]X^-|X,E^@_(M+//M)=0.44 V and E_(X//X^-)^@ =0.33 V . From this data, one can duduce that :

The lowering of vapour pressure of 0.1M aqueous solutions of NaCl, CuSO_4 , and K_2SO_4 are:

A strip of nickel metal is dipped in a 1 molar solution of Ni (NO_(3))_(2) and a strip of silver metal is dipped in a 1 molar solution of AgNO_(3) . An electrochemical cell is created when the two solutions are joined by salt bridge and two strips are joined by wire to a voltmeter. Answer the following questions. Calculate the cell potential (E_("cell")) at 25^(@)C for the cell if the initial concentration of Ni (NO_(3))_(2) , is 0.100 molar and the initial concentration of AgNO_(3) is 1.00 molar. [E_(Ni^(2+)//Ni)= -0.25V, E_("Ag^(+)//Ag)=0.80V, log 10^(-1)= -1]